I have DB2 table like below -
Date1 Item_code Amt
2018-06-01 1 2
2018-06-02 1 3
2018-06-03 2 4
2018-06-03 2 5
2018-06-04 3 6
2018-06-05 3 7
2018-06-06 4 8
I need the cumulative sum item_code wise per day. The result should look like -
Date1 Item_code Amt
2018-06-01 1 2
2018-06-02 1 5
2018-06-03 2 9
2018-06-04 3 6
2018-06-05 3 13
2018-06-06 4 8
I have tried a lot by myself and search also on SO but nothing is fulfilling my need. There are a lot of examples if I just need the cumulative sum day wise irrespective of item code.
Any help is greatly appreciated. Thanks in advance.
I think you want aggregation with a cumulative sum:
select item_code, date1,
sum(sum(amt)) over (partition by item_code order by date1) as running_amt
from t
group by item_code, date;
Related
I need to show the number of valid inspectors we have by month over the last five years. Inspectors are considered valid when the expiration date on their certification has not yet passed, recorded as the month end date. The below SQL code is text of the query to count valid inspectors for January 2017:
SELECT Count(*) AS RecordCount
FROM dbo_Insp_Type
WHERE (dbo_Insp_Type.CERT_EXP_DTE)>=#2/1/2017#);
Rather than designing 60 queries, one for each month, and compiling the results in a final table (or, err, query) are there other methods I can use that call for less manual input?
From this sample:
Id
CERT_EXP_DTE
1
2022-01-15
2
2022-01-23
3
2022-02-01
4
2022-02-03
5
2022-05-01
6
2022-06-06
7
2022-06-07
8
2022-07-21
9
2022-02-20
10
2021-11-05
11
2021-12-01
12
2021-12-24
this single query:
SELECT
Format([CERT_EXP_DTE],"yyyy/mm") AS YearMonth,
Count(*) AS AllInspectors,
Sum(Abs([CERT_EXP_DTE] >= DateSerial(Year([CERT_EXP_DTE]), Month([CERT_EXP_DTE]), 2))) AS ValidInspectors
FROM
dbo_Insp_Type
GROUP BY
Format([CERT_EXP_DTE],"yyyy/mm");
will return:
YearMonth
AllInspectors
ValidInspectors
2021-11
1
1
2021-12
2
1
2022-01
2
2
2022-02
3
2
2022-05
1
0
2022-06
2
2
2022-07
1
1
ID
Cert_Iss_Dte
Cert_Exp_Dte
1
1/15/2020
1/15/2022
2
1/23/2020
1/23/2022
3
2/1/2020
2/1/2022
4
2/3/2020
2/3/2022
5
5/1/2020
5/1/2022
6
6/6/2020
6/6/2022
7
6/7/2020
6/7/2022
8
7/21/2020
7/21/2022
9
2/20/2020
2/20/2022
10
11/5/2021
11/5/2023
11
12/1/2021
12/1/2023
12
12/24/2021
12/24/2023
A UNION query could calculate a record for each of 50 months but since you want 60, UNION is out.
Or a query with 60 calculated fields using IIf() and Count() referencing a textbox on form for start date:
SELECT Count(IIf(CERT_EXP_DTE>=Forms!formname!tbxDate,1,Null)) AS Dt1,
Count(IIf(CERT_EXP_DTE>=DateAdd("m",1,Forms!formname!tbxDate),1,Null) AS Dt2,
...
FROM dbo_Insp_Type
Using the above data, following is output for Feb and Mar 2022. I did a test with Cert_Iss_Dte included in criteria and it did not make a difference for this sample data.
Dt1
Dt2
10
8
Or a report with 60 textboxes and each calls a DCount() expression with criteria same as used in query.
Or a VBA procedure that writes data to a 'temp' table.
I am looking to filter very large tables to the latest entry per user per month. I'm not sure if I found the best way to do this. I know I "should" trust the SQL engine (snowflake) but there is a part of me that does not like the join on three columns.
Note that this is a very common operation on many big tables, and I want to use it in DBT views which means it will get run all the time.
To illustrate, my data is of this form:
mytable
userId
loginDate
year
month
value
1
2021-01-04
2021
1
41.1
1
2021-01-06
2021
1
411.1
1
2021-01-25
2021
1
251.1
2
2021-01-05
2021
1
4369
2
2021-02-06
2021
2
32
2
2021-02-14
2021
2
731
3
2021-01-20
2021
1
258
3
2021-02-19
2021
2
4251
3
2021-03-15
2021
3
171
And I'm trying to use SQL to get the last value (by loginDate) for each month.
I'm currently doing a groupby & a join as follows:
WITH latest_entry_by_month AS (
SELECT "userId", "year", "month", max("loginDate") AS "loginDate"
FROM mytable
)
SELECT * FROM mytable NATURAL JOIN latest_entry_by_month
The above results in my desired output:
userId
loginDate
year
month
value
1
2021-01-25
2021
1
251.1
2
2021-01-05
2021
1
4369
2
2021-02-14
2021
2
731
3
2021-01-20
2021
1
258
3
2021-02-19
2021
2
4251
3
2021-03-15
2021
3
171
But I'm not sure if it's optimal.
Any guidance on how to do this faster? Note that I am not materializing the underlying data, so it is effectively un-clustered (I'm getting it from a vendor via the Snowflake marketplace).
Using QUALIFY and windowed function(ROW_NUMBER):
SELECT *
FROM mytable
QUALIFY ROW_NUMBER() OVER(PARTITION BY userId, year, month
ORDER BY loginDate DESC) = 1
Is there a way to find the solution so that I need for 2 days, there are 2 UD's because there are June 24 2 times and for the rest there are single days.
I am showing the expected output here:
Primary key UD Date
-------------------------------------------
1 123 2015-06-24 00:00:00.000
6 456 2015-06-24 00:00:00.000
2 123 2015-06-25 00:00:00.000
3 658 2015-06-26 00:00:00.000
4 598 2015-06-27 00:00:00.000
5 156 2015-06-28 00:00:00.000
No of times Number of days
-----------------------------
4 1
2 2
The logic is 4 users are there who used the application on 1 day and there are 2 userd who used the application on 2 days
You can use two levels of aggregation:
select cnt, count(*)
from (select date, count(*) as cnt
from t
group by date
) d
group by cnt
order by cnt desc;
I have a table with 1 result per day like this :
id | item_id | date | amount
-------------------------------------
1 1 2019-01-01 1
2 1 2019-01-02 2
3 1 2019-01-03 3
4 1 2019-01-04 4
5 1 2019-01-05 5
6 2 2019-01-01 1
7 2 2019-01-01 2
8 2 2019-01-01 3
9 2 2019-01-01 4
10 2 2019-01-01 5
11 3 2019-01-01 1
12 3 2019-01-01 2
13 3 2019-01-01 3
14 3 2019-01-01 4
15 3 2019-01-01 5
First I was trying to average the column amount for each day.
SELECT
x.item_id AS id,avg(x.amount) AS result
FROM
(SELECT
il.item_id, il.amount,
ROW_NUMBER() OVER (PARTITION BY il.item_id ORDER BY il.date DESC) rn
FROM
item_prices il) x
WHERE
x.rn BETWEEN 1 AND 50
GROUP BY
x.item_id
The result is going to be the following if calculated on 2019-01-05
item_id | average
1 3
2 3
3 3
or, if calculated 2019-01-04
item_id | average
1 2.5
2 2.5
3 2.5
My goal is to run the Average query , every day that would update the average automatically and insert it in 5th column "average" :
id | item_id | date | amount | average
5 1 2019-01-05 5 3
10 2 2019-01-05 5 3
15 3 2019-01-05 5 3
Issue is that every example i can find with Insert the Select they only update one row and they are over another table there is also the most recent date issue...
Can someone point me in the right direction?
Perhaps you want to see running average every day. Storing the value as a separate column is bound to cause problems especially when the rows are updated/deleted, the column also needs to be updated and hence will require complex triggers.
Simply create a View and run whenever you want to check the average directly from that View.
CREATE OR REPLACE VIEW v_item_prices AS
SELECT t.*,avg(t.amount) OVER ( PARTITION BY item_id order by date)
AS average FROM item_prices t
order by item_id,date
DEMO
I have a database with the following data:
Group ID Time
1 1 16:00:00
1 2 16:02:00
1 3 16:03:00
2 4 16:09:00
2 5 16:10:00
2 6 16:14:00
I am trying to find the difference in times between the consecutive rows within each group. Using LAG() and DATEDIFF() (ie. https://stackoverflow.com/a/43055820), right now I have the following result set:
Group ID Difference
1 1 NULL
1 2 00:02:00
1 3 00:01:00
2 4 00:06:00
2 5 00:01:00
2 6 00:04:00
However I need the difference to reset when a new group is reached, as in below. Can anyone advise?
Group ID Difference
1 1 NULL
1 2 00:02:00
1 3 00:01:00
2 4 NULL
2 5 00:01:00
2 6 00:04:00
The code would look something like:
select t.*,
datediff(second, lag(time) over (partition by group order by id), time)
from t;
This returns the difference as a number of seconds, but you seem to know how to convert that to a time representation. You also seem to know that group is not acceptable as a column name, because it is a SQL keyword.
Based on the question, you have put group in the order by clause of the lag(), not the partition by.